# Thread: Vector space

1. ## Vector space

I don't really understand some of this notation. I was wondering if someone could help me determine which of the following are subspaces and why they fail to be subspaces (if they are not).

Let R[X] denote the vector space (over R) of polynomials with real coefficients. Which of the following are subspaces of R[X]?

a) The set of odd polynomials.
b) The set of polynomials of odd degree (We include the polynomial 0).
c) {p(X) in R[X]: p(i*sqrt(3))=0}.
d) {p(X) in R[X]: p(0)= i*sqrt(3)}.
e) The set {(X^2+3)(p(X)): p(X) in R[X]}.
f) The set {(X-i*sqrt(3))p(X): p(X) in R[X]}.
g) The set of polynomials with rational coefficients.
h) {p(X) in R[X]: p(i*sqrt(3)) is real}.

Thank you so much for any help.

2. Originally Posted by thesummerofgeorge
I don't really understand some of this notation. I was wondering if someone could help me determine which of the following are subspaces and why they fail to be subspaces (if they are not).

Let R[X] denote the vector space (over R) of polynomials with real coefficients. Which of the following are subspaces of R[X]?
The fundamental theorem states this: if $V$ is a vector space over the real field $\mathbb{R}$ and some subset of $V$ is closed under vector addition and scalar multiplication that is:
$\alpha+\beta \in V$
$a\alpha \in V$
For all vectos in the subset and all real scalars then this subset is in fact a subspace.

a) The set of odd polynomials.
And odd polynomial, meaning
$P(-x)=-P(x)$
Thus,
$P(x)+Q(x)$
$P(-x)+Q(-x)=-P(x)-Q(x)=-(P(x)+Q(x))$
Thus it is closed under vector addition.
$kP(x)$
$kP(-x)=-kP(x)$
Thus, it is closer under scalar multiplication.

3. Originally Posted by thesummerofgeorge
I don't really understand some of this notation. I was wondering if someone could help me determine which of the following are subspaces and why they fail to be subspaces (if they are not).

Let R[X] denote the vector space (over R) of polynomials with real coefficients. Which of the following are subspaces of R[X]?

a) The set of odd polynomials.
b) The set of polynomials of odd degree (We include the polynomial 0).
Not a sumspace as x^3, -x^3+x^2 are both in the set, but their sum
is not.

c) {p(X) in R[X]: p(i*sqrt(3))=0}.
Is a subspace, as if P and Q are in this set, for any a and b in R, we
have:

aP(i*sqrt(3)) + bQ(i*sqrt(3))=a*0+b*0=0.

(also the set is not empty, but if it were it would still be a subspace)

d) {p(X) in R[X]: p(0)= i*sqrt(3)}.
If the set is non-empty then this is not, using the notation of part c) above, and a != -b

aP(0)+bQ(0)=(a+b)*i*sqrt(3) != i*sqrt(3)

however the set is empty, as any polynomial in x for which p(0)=i*sqrt(3)
is not a polynomial with real coeficients, Hence this set is the null set and so
is a sub-space

RonL

4. Originally Posted by thesummerofgeorge
e) The set {(X^2+3)(p(X)): p(X) in R[X]}.
First the set is not empty. Let P(x)=(x^2+3)p(x), for some p(x) in R[x],
and Q(x)=(x^2+3)q(x), for some q(x) in R[x], then these are both in R[x],
and for any a and b in R:

aP(x)+bQ(x)=(x^2+3)(p(x)+q(x))

hence is in the set, so the set is a subspace.

f) The set {(X-i*sqrt(3))p(X): p(X) in R[X]}.
Is not a subspace as the elements of this set are not polynomials
in x with real coeficients.

g) The set of polynomials with rational coefficients.
h) {p(X) in R[X]: p(i*sqrt(3)) is real}.
Both of these are but I will leave it to you or some other helper
to show they are.

RonL